To solve the give logarithm we proceed as follows; log7(3x^3+x)-log7(x)=2 this can be written as: log7[(3x^3+x)/(x)]=2
but; log7(49)=2 thus; log7(3x^3+x)/x=log7(49) hence canceling log7 we get: (3x^3+x)/x=49 This can be simplified further to get: 3x^2+1=49 3x^2=49-1 3x^2=48 x^2=48/3 x^2=16 getting the square root of both sides we get: x=4 the answer is x=4
